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EAS 6939 Aerospace Structural Composites
Solution for Project Phase-I
Prepared By:
Sameer Luthra and Raja Ganesh
Course Advisor
Dr. Raphael T. Haftka
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Table of Contents
Contents
Inputs ............................................................................................................................................................ 3
Steps Followed and Formulas Used .............................................................................................................. 4
Part 1(a): Using 1 Pair of orientation angles ±𝜽........................................................................................... 8
Part 1(b): Using 2 Pairs of orientation angles ±𝜽𝟏, ±𝜽𝟐 ............................................................................. 9
Part 2: Minimizing the Poisson’s ratio ........................................................................................................ 10
Part 3: Lightest laminate of the form 𝟎𝒏𝟏/±𝟒𝟓𝒏𝟐/𝟗𝟎𝒏𝟑𝒔...................................................................... 11
Appendix 1: Excel File with answer reports ................................................................................................ 12
Appendix 2: Grader’s Feedback .................................................................................................................. 12
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Inputs
A Cylindrical Pressure vessel of 𝑹𝒂𝒅𝒊𝒖𝒔 = 𝟐𝒎 is to be designed for 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 = 𝟎. 𝟖 𝑴𝑷𝒂.
Material Used: Graphite-Epoxy
Material Properties:
𝑋𝑡 = 𝑋𝑐 = 1500𝑀𝑃𝑎
𝑌𝑡 = 40𝑀𝑃𝑎
𝑌𝑐 = 246𝑀𝑃𝑎
𝑆 = 68𝑀𝑃𝑎
𝐸1 = 181𝐺𝑃𝑎
𝐸2 = 10.3𝐺𝑃𝑎
𝐺12 = 7.17𝐺𝑃𝑎
𝜈12 = 0.28
𝛼1 = 0.02 × 10−6 /℃
𝛼2 = 22.5 × 10−6 /℃
Other Inputs:
𝑡𝑝𝑙𝑦 = 0.125𝑚𝑚
𝑇𝑐𝑢𝑟𝑒 = 180℃
𝑇𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 = 10℃
∆𝑇 = 𝑇𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 − 𝑇𝑐𝑢𝑟𝑒 = −170℃
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Steps Followed and Formulas Used
 Minor Poisson’s Ratio for Ply is calculated as:
𝜈21 =
𝐸2
𝜈
𝐸1 12
 Matrix [Q] is calculated:
Q11
Matrix [Q] = [Q12
0
Q12
Q22
0
0
0 ]
Q66
Where
𝑄11 =
𝐸1
,
1 − 𝜈12 𝜈21
𝑄12 =
𝑄22 =
𝐸12
1 − 𝜈12 𝜈21
𝜈12 𝐸2
𝜈21 𝐸1
=
1 − 𝜈12 𝜈21 1 − 𝜈12 𝜈21
𝑄66 = 𝐺12
 Material Invariants U’s are then calculated as:
1
𝑈1 = (3𝑄11 + 3𝑄22 + 2𝑄12 + 4𝑄66 )
8
1
𝑈2 = 2 (𝑄11 − 𝑄22 )
𝑈3 =
1
(𝑄 + 𝑄22 − 2𝑄12 − 4𝑄66 )
8 11
𝑈4 =
1
(𝑄 + 𝑄22 + 6𝑄12 − 4𝑄66 )
8 11
𝑈5 =
1
(𝑄 + 𝑄22 − 2𝑄12 + 4𝑄66 )
8 11
 Lamination Parameters are then calculated as:
𝐼
𝑉1∗
= ∑ 𝜈𝑘 cos 2𝜃𝑘
𝑘=1
𝐼
𝑉3∗
= ∑ 𝜈𝑘 cos 4𝜃𝑘
𝑘=1
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𝑤ℎ𝑒𝑟𝑒 𝐼 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 ± 𝜃 𝑔𝑟𝑜𝑢𝑝𝑠
𝜈𝑖 : 𝑉𝑜𝑙𝑢𝑚𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑦𝑒𝑟𝑠 𝑤𝑖𝑡ℎ ± 𝜃𝑖 𝑜𝑟𝑖𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑔𝑙𝑒𝑠
𝜈𝑖 =
𝑁𝑜. 𝑜𝑓 𝐿𝑎𝑦𝑒𝑟𝑠 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 ± 𝜃𝑖 𝑜𝑟𝑖𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑔𝑙𝑒
𝑇𝑜𝑡𝑎𝑙 𝑁𝑜. 𝑜𝑓 𝐿𝑎𝑦𝑒𝑟𝑠
 Matrix [𝐴∗ ] is calculated:
A∗11 A∗12
0
∗
∗
A
A
0 ]
= [ 12
22
0
0 A∗66
∗
𝑤ℎ𝑒𝑟𝑒 𝐴11
= 𝑈1 + 𝑉1∗ 𝑈2 + 𝑉3∗ 𝑈3
[𝐴∗ ]
𝐴∗22 = 𝑈1 − 𝑉1∗ 𝑈2 + 𝑉3∗ 𝑈3
∗
𝐴12
= 𝑈4 − 𝑉3∗ 𝑈3
𝐴∗66 = 𝑈5 − 𝑉3∗ 𝑈3
 Some initial value of Laminate thickness, h is assumed which is iterated to find the optimal
value.
 Stresses in the Laminate due to Internal Pressure are calculated as:
𝜎𝑥𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 × 𝑅𝑎𝑑𝑖𝑢𝑠
2 × 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
(𝑎𝑥𝑖𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑠𝑡𝑟𝑒𝑠𝑠)
𝜎𝑦𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 × 𝑅𝑎𝑑𝑖𝑢𝑠
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
(ℎ𝑜𝑜𝑝 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑠𝑡𝑟𝑒𝑠𝑠)
 Using these Stresses in the laminate due to Internal Pressure, the Laminate mid-plane
Strains(Due to Pressure) for the Balanced Symmetric laminate are found as:
(𝜀𝑥0 )𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝜎𝑥𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
0
∗
−1
{ (𝜀𝑦 )𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 } = [𝐴 ] {𝜎 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 }
𝑦
0
(𝜀𝑥𝑦
)𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
0
 To find the average Thermal Loads in the Laminate:
1
𝑁𝑥𝑇ℎ𝑒𝑟𝑚𝑎𝑙 = ∗ ℎ ∗ (𝐾1 + 𝐾2 𝑉1∗ )Δ𝑇
2
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𝑁𝑦𝑇ℎ𝑒𝑟𝑚𝑎𝑙 =
𝑇ℎ𝑒𝑟𝑚𝑎𝑙
𝑁𝑥𝑦
=
1
∗ ℎ ∗ (𝐾1 − 𝐾2 𝑉1∗ )Δ𝑇
2
1
∗ ℎ ∗ (𝐾3 𝑉2∗ )Δ𝑇 = 0 (𝑉2∗ = 0 𝐹𝑜𝑟 𝑎 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝐵𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑙𝑎𝑚𝑖𝑛𝑎𝑡𝑒)
2
𝑤ℎ𝑒𝑟𝑒:
𝐾1 = (𝑈1 + 𝑈4 )(𝛼1 + 𝛼2 ) + 𝑈2 (𝛼1 − 𝛼2 )
& 𝐾2 = 𝑈2 (𝛼1 + 𝛼2 ) + (𝑈1 + 2𝑈3 − 𝑈4 )(𝛼1 − 𝛼2 )
 Using these Thermal Loads in the laminate, the Laminate Non-Mechanical Strains for the
Balanced Symmetric laminate are found as:
𝑁𝑥𝑇ℎ𝑒𝑟𝑚𝑎𝑙
𝜀𝑥𝑜𝑁
1
{𝜀𝑦𝑜𝑁 } = ∗ [𝐴∗ ]−1 {𝑁𝑦𝑇ℎ𝑒𝑟𝑚𝑎𝑙 }
ℎ
𝑜𝑁
𝑇ℎ𝑒𝑟𝑚𝑎𝑙
𝜀𝑥𝑦
𝑁𝑥𝑦
 To transform strains in a Laminate to strains in a ply with ±𝜃𝑖 orientation angle:
𝜃
𝜀1 𝑖
𝑚2
𝜃𝑖
{ 𝜀2 } = [ 𝑛2
𝜃
−2𝑚𝑛
𝛾 𝑖
12
𝑛2
𝑚2
2𝑚𝑛
𝑤ℎ𝑒𝑟𝑒: 𝑚 = cos 𝜃𝑖
𝜀𝑥0
2𝑚𝑛
0
−2𝑚𝑛 ] { 𝜀𝑦 }
0
𝑚2 − 𝑛2 𝜀𝑥𝑦
𝑎𝑛𝑑
𝑛 = sin 𝜃𝑖
 To find Free strains in Laminate (x-y) co-ordinate system in a ply with ±𝜃𝑖 orientation
angle:
(𝜃 )
𝑖
𝜀𝑥𝐹
𝛼1 𝑚2 + 𝛼2 𝑛2
𝐹
{ 𝜀𝑦 }
= { 𝛼1 𝑛2 + 𝛼2 𝑚2 } Δ𝑇
𝐹
2 ∗ (𝛼1 − 𝛼2 )𝑚𝑛
𝛾𝑥𝑦
𝑤ℎ𝑒𝑟𝑒: 𝑚 = cos 𝜃𝑖 𝑎𝑛𝑑 𝑛 = sin 𝜃𝑖
 So the Net thermal strains in Laminate (x-y) co-ordinate system in a ply with ±𝜃𝑖
orientation angle:
(𝜃𝑖 )
𝜀𝑥𝑅
{ 𝜀𝑦𝑅 }
𝑅
𝛾𝑥𝑦
(𝜃𝑖 )
𝜀𝑥𝑜𝑁
𝜀𝑥𝐹
= {𝜀𝑦𝑜𝑁 } − { 𝜀𝑦𝐹 }
𝑜𝑁
𝐹
𝜀𝑥𝑦
𝛾𝑥𝑦
 So the Effective Strains due to Internal Pressure & due to change in temperature ΔT in
Laminate (x-y) co-ordinate system in a ply with ±𝜃𝑖 orientation angle:
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(𝜃𝑖 )
𝐸𝑓𝑓
𝜀𝑥
𝐸𝑓𝑓
{ 𝜀𝑦
}
𝐸𝑓𝑓
𝛾𝑥𝑦
(𝜃𝑖 )
(𝜀𝑥0 )𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝜀𝑥𝑅
0
= { (𝜀𝑦 )𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 } + { 𝜀𝑦𝑅 }
𝑅
0
𝛾𝑥𝑦
(𝜀𝑥𝑦
)
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
 To transform Effective strains from Laminate (x-y) co-ordinate system to Ply (1-2) coordinate system in a ply with ±𝜃𝑖 orientation angle:
𝜃
𝜀1 𝑖
𝑚2
𝜃𝑖
{ 𝜀2 } = [ 𝑛2
𝜃
−2𝑚𝑛
𝛾 𝑖
12
𝑛2
𝑚2
2𝑚𝑛
(𝜃𝑖 )
𝜀𝑥𝑁𝑒𝑡
2𝑚𝑛
𝑁𝑒𝑡
−2𝑚𝑛 ] { 𝜀𝑦 }
𝑁𝑒𝑡
𝑚2 − 𝑛2 𝛾𝑥𝑦
𝑤ℎ𝑒𝑟𝑒: 𝑚 = cos 𝜃𝑖
𝑎𝑛𝑑
𝑛 = sin 𝜃𝑖
 To find stresses in Ply (1-2) co-ordinate system in a ply with ±𝜃𝑖 orientation angle:
𝜃
𝜎1 𝑖
𝜃
𝜀1 𝑖
{𝜎2𝜃𝑖 } = [𝑄] { 𝜀2𝜃𝑖 }
𝜃
𝜏12𝑖
𝜃
𝛾12𝑖
 Hoffman Criterion is applied separately to each ±𝛉𝐢 pair of plies to design the lightest
balanced and symmetric laminate:
𝜃
𝜃
𝜃
𝜃
𝜃
(𝜎1 𝑖 )2 (𝜎1 𝑖 )(𝜎2 𝑖 ) (𝜎2 𝑖 )2
1
1
1 1
(𝜏 𝑖 )2
𝜃
𝜃
−
+
+ ( − ) (𝜎1 𝑖 ) + ( + ) (𝜎2 𝑖 ) + 122 ≤ 1
𝑋𝑡 𝑋𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑆
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Part 1(a): Using 1 Pair of orientation angles ±𝜽
Objective: Design of the lightest Balanced and Symmetric laminate that can carry safely all
the loads. In this part we want to design the lightest angle-ply laminate [(±𝜽)𝒏 ]𝒔 (where n is a
continuous variable) based on Hoffman Criterion.
Procedure to solve: First of all the design problem is formulated as:
Objective Function: Minimize the Thickness of the laminate i.e.
𝒎𝒊𝒏𝒊𝒎𝒊𝒛𝒆
ℎ
Design Variables: 𝜃 & 𝑛
Constraints: The design is based on the Hoffman Criterion:
2
𝜎12
𝜎1 𝜎2
𝜎22
1
1
1 1
𝜏12
−
+
+ ( − ) 𝜎1 + ( + ) 𝜎2 + 2 ≤ 1
𝑋𝑡 𝑋𝑐 𝑋𝑡 𝑋𝑐 𝑌𝑡 𝑌𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑆
Steps Followed: Formulas used and general steps followed have been discussed in the initial
section. Initial values of 𝜃 & n are taken:
𝜃𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 50°
𝑛𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 1
These values are then iterated to find the optimal solution.
Finally, n is rounded up to an integer value.
Results: The optimal values are:
𝒉 = 𝟏𝟔. 𝟓 𝒎𝒎
𝜽 = 𝟓𝟗. 𝟒°
𝐧 =33
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 = 𝟏𝟑𝟐
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Part 1(b): Using 2 Pairs of orientation angles ±𝜽𝟏 , ±𝜽𝟐
Objective: Design of the lightest Balanced and Symmetric laminate that can carry safely all
the loads. In this part we want to design the lightest laminate with 2 pair of orientation angles
[(±𝜽𝟏 )𝒏𝟏 /(±𝜽𝟐 )𝒏𝟐 ]𝒔 (where 𝒏𝟏 & 𝒏𝟐 are a continuous variables) based on Hoffman
Criterion.
Procedure to solve: First of all the design problem is formulated as:
Objective Function: Minimize the Thickness of the laminate i.e.
𝒎𝒊𝒏𝒊𝒎𝒊𝒛𝒆
ℎ
Design Variables: 𝜃1 , 𝜃2 , 𝑛1 , 𝑛2
Constraints: The design is based on the Hoffman Criterion:
𝜃
𝜃
𝜃
𝜃
𝜃
(𝜎1 𝑖 )2 (𝜎1 𝑖 )(𝜎2 𝑖 ) (𝜎2 𝑖 )2
1
1
1 1
(𝜏12𝑖 )2
𝜃𝑖
𝜃𝑖
−
+
+ ( − ) (𝜎1 ) + ( + ) (𝜎2 ) +
≤1
𝑋𝑡 𝑋𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑆2
𝑖 = 1,2
Steps Followed: Formulas used and general steps followed have been discussed in the initial
section. Major steps followed till finding Laminate mid-plane strains are the same as the
previous part.
 Initial values of 𝜃1 , 𝜃2 , 𝑛1 , 𝑛2 are taken:
(𝜃1 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 50°
(𝜃2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 50°
(𝑛1 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 1
(𝑛2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 1
These values are then iterated to find the optimal solution.
𝐧𝟏 𝐢𝐬 𝐫𝐨𝐮𝐧𝐝𝐞𝐝 𝐝𝐨𝐰𝐧 𝐚𝐧𝐝 𝐧𝟐 𝐢𝐬 𝐫𝐨𝐮𝐧𝐝𝐞𝐝 𝐮𝐩 since the laminate thickness is still greater than the
optimal thickness when n1 and n2 were considered as continuous variables
Results: The optimal values are:
𝒉 = 𝟏𝟔. 𝟓𝒎𝒎
𝜽𝟏 = 𝟓𝟗. 𝟒°
𝜽𝟐 = 𝟓𝟗. 𝟒°
𝒏𝟏 = 𝟏𝟔
𝒏𝟐 = 𝟏𝟕
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 = 𝟒 ∗ (𝒏𝟏 + 𝒏𝟐 ) = 𝟏𝟑𝟐
Note: This is the same laminate that was obtained in section 1-a.
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Part 2: Minimizing the Poisson’s ratio
Objective: Using the Balanced Symmetric laminate [(±𝜃)𝑛 ]𝑠 (with n = integer) designed in
the previous part, design a Balanced Symmetric laminate with the same thickness (as calculated
in part 1) that will not fail and has the lowest Poisson’s Ratio(considering the maximum of the
two Poisson’s ratios)
Procedure to solve: First of all the design problem is formulated as:
Objective Function: Minimize the Poisson’s Ratio (Maximum of the two Poisson’s ratios).
𝒎𝒊𝒏𝒊𝒎𝒊𝒛𝒆
max(𝜈𝑥𝑦 , 𝜈𝑦𝑥 )
Design Variable: 𝑂𝑟𝑖𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑔𝑙𝑒 𝜃
Constraints: Constraints: The design is based on the Hoffman Criterion:
2
𝜎12
𝜎1 𝜎2
𝜎22
1
1
1 1
𝜏12
−
+
+ ( − ) 𝜎1 + ( + ) 𝜎2 + 2 ≤ 1
𝑋𝑡 𝑋𝑐 𝑋𝑡 𝑋𝑐 𝑌𝑡 𝑌𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑆
Steps Followed: Most of the Formulas used and general steps followed have been discussed in
the initial section. The n value (rounded to integer) obtained from Part 1 is kept constant.
 Formulas for Poisson’s Ratios of the Laminate:
∗
𝐴12
𝜈𝑥𝑦 = ∗
𝐴22
𝐸𝑦
𝜈𝑦𝑥 =
𝜈
𝐸𝑥 𝑥𝑦
 The value of Orientation angle θ is then iterated to find the optimal solution (i.e. to
Minimize(max(𝛎𝐱𝐲 , 𝛎𝐲𝐱 )))
Results: The final values are:
𝒉 = 𝟏𝟔. 𝟓 𝒎𝒎
𝜽 = 𝟓𝟖. 𝟐𝟓°
𝒏 = 𝟑𝟑
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 = 𝟏𝟑𝟐
𝝂𝒙𝒚 = 𝟎. 𝟑𝟑𝟓
𝝂𝒚𝒙 = 𝟏. 𝟑𝟏𝟑
11 | P a g e
Part 3: Lightest laminate of the form [𝟎𝒏𝟏 /(±𝟒𝟓)𝒏𝟐 /𝟗𝟎𝒏𝟑 ]
𝒔
Objective: Design of the lightest Balanced and Symmetric laminate that can carry safely all
the loads. The laminate is designed based on Hoffman Criterion. In this part we want to design
the laminate [𝟎𝒏𝟏 /(±𝟒𝟓)𝒏𝟐 /𝟗𝟎𝒏𝟑 ]𝒔 with Minimum thickness.
Procedure to solve: First of all the design problem is formulated as:
Objective Function: Minimize the Thickness of the laminate i.e.
𝒎𝒊𝒏𝒊𝒎𝒊𝒛𝒆
ℎ
Design Variables: 𝑛1 , 𝑛2 , 𝑛3
Constraints: The design is based on the Hoffman Criterion. There is another constraint that
volume fraction of any orientation angle should not be less than 0.10 and should not be greater
than 0.50.
𝜃
𝜃
𝜃
𝜃
𝜃
(𝜎1 𝑖 )2 (𝜎1 𝑖 )(𝜎2 𝑖 ) (𝜎2 𝑖 )2
1
1
1 1
(𝜏12𝑖 )2
𝜃𝑖
𝜃𝑖
−
+
+ ( − ) (𝜎1 ) + ( + ) (𝜎2 ) +
≤1
𝑋𝑡 𝑋𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑋𝑡 𝑋𝑐
𝑌𝑡 𝑌𝑐
𝑆2
𝑤ℎ𝑒𝑟𝑒 𝜃1 = 00 ,
𝜃2 = 450 ,
𝑖 = 1,2,3
𝜃3 = 900
0.10 ≤ (𝑉𝑜𝑙𝑢𝑚𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛)0° ≤ 0.50
0.10 ≤ (𝑉𝑜𝑙𝑢𝑚𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛)±45° ≤ 0.50
0.10 ≤ (𝑉𝑜𝑙𝑢𝑚𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛)90° ≤ 0.50
Steps Followed: Formulas used and general steps followed have been discussed in the initial
section. Some of the major steps followed are summarized as:
 Initial values of 𝑛1 , 𝑛2 , 𝑛3 are taken:
𝑛1 = 1
𝑛2 = 1
𝑛3 = 1
These values are then iterated to find the optimal solution.
𝐧𝟏 𝐢𝐬 𝐫𝐨𝐮𝐧𝐝𝐞𝐝 𝐝𝐨𝐰𝐧 𝐚𝐧𝐝 𝐧𝟐 𝐚𝐧𝐝 𝐧𝟑 𝐚𝐫𝐞 𝐫𝐨𝐮𝐧𝐝𝐞𝐝 𝐮𝐩 since the laminate thickness is still greater
than the optimal thickness when n1 , n2 and n3 were considered as continuous variables
12 | P a g e
The volume fractions of plies of each orientation are computed as follows:
𝑛1
𝑛1 + 2𝑛2 + 𝑛3
(𝑉𝑜𝑙𝑢𝑚𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛)0° =
(𝑉𝑜𝑙𝑢𝑚𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛)45° =
2𝑛2
(𝑛1 + 2𝑛2 + 𝑛3 )
(𝑉𝑜𝑙𝑢𝑚𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛)90° =
𝑛3
(𝑛1 + 2𝑛2 + 𝑛3 )
Results: The optimal values are:
𝒉 = 𝟒𝟖. 𝟕𝟓 𝒎𝒎
𝒏𝟏 = 𝟗𝟕
𝒏𝟐 = 𝟑𝟗
𝒏𝟑 = 𝟐𝟎
𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 𝒐𝒇 𝟎° = 𝟗𝟕
𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 𝒐𝒇 ± 𝟒𝟓° = 𝟑𝟗
𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 𝒐𝒇 𝟗𝟎° = 𝟐𝟎
𝑻𝒐𝒕𝒂𝒍 𝑵𝒐. 𝒐𝒇 𝑷𝒍𝒊𝒆𝒔 = 𝟐 ∗ (𝒏𝟏 + 𝟐𝒏𝟐 + 𝒏𝟑 ) = 𝟑𝟗𝟎
Appendix 1: Excel File with answer reports
Project_Phase1_Solu
tion_Raja.xlsx
Double-click to open
Appendix 2: Grader’s Feedback
Project_Phase1_Fee
dback.pdf
Double-click to open